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Strike441 [17]
3 years ago
12

What is the charge of a particle having 7 protons and 10 electrons?

Chemistry
1 answer:
kirill115 [55]3 years ago
3 0

Answer:

Nitrogen has an atomic number of 7; the neutral Nitrogen atom has 7 protons and 7 electrons. If Nitrogen gained three electrons it would have 10 electrons, like the Noble gas Neon (10 protons, 10 electrons). However, unlike Neon, the resulting Nitrogen ion would have a net charge of N3- (7 protons, 10 electrons).

Explanation:

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467.5 centimeters. are you sure that's high school chemistry though, lad? 
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1. A student wants to test if giving dogs a toy to play with makes their tail wag more. He gives dog toys to one group, while an
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4 years ago
Calculate the mass percent (m/m) of a solution prepared by dissolving 45.09 g of NaCl in 174.9 g of H2O.
inysia [295]

Answer:

20.3 %  NaCl

Explanation:

Given data:

Mass of solute = 45.09 g

Mass of solvent = 174.9 g

Mass percent of solution = ?

Solution:

Mass of solution = 45.09 g + 174.9 g

Mass of solution = 220 g

The solute in 220 g is 45.09 g

220 g = 2.22 × 45.09

In 100 g solution amount of solute:

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5 0
3 years ago
Preparation of Standard Buffer for Calibration of a pH Meter The glass electrode used in commercial pH meters gives an electrica
lana66690 [7]

Answer:

  • Mass of NaH₂PO₄·H₂O = 8.542 g
  • Mass of Na₂HPO₄ = 5.410 g

Explanation:

Keeping in mind the equilibrium:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺

We use the Henderson-Hasselbalch equation (H-H):

pH = pka + log\frac{[A^{-}]}{[HA]}

For this problem [A⁻] = [HPO₄⁻²] and [HA] = [H₂PO₄⁻]

From literature we know that pka = 7.21, from the problem we know that pH=7.00 and that

[HPO₄⁻²] + [H₂PO₄⁻] = 0.100 M

From this equation we can <u>express [H₂PO₄⁻] in terms of [HPO₄⁻²]</u>:

[H₂PO₄⁻] = 0.100 M - [HPO₄⁻²]

And then replace [H₂PO₄⁻] in the H-H equation, <u>in order to calculate [HPO₄⁻²]</u>:

7.00=7.21+log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]} \\-0.21=log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\10^{-0.21} =\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\0.616*(0.100M-[HPO4^{-2}])=[HPO4^{-2}]\\0.0616 M = 1.616*[HPO4^{-2}]\\0.03812 M =[HPO4^{-2}]

With the value of  [H₂PO₄⁻],<u> we calculate [HPO₄⁻²]</u>:

[HPO₄⁻²] + 0.0381 M = 0.100 M

[HPO₄⁻²] = 0.0619 M

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  • Mass of NaH₂PO₄·H₂O = 0.0619 M * 1 L * 138 g/mol = 8.542 g
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8 0
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